Olympiad problems

1999 All-Russian Olympiad Regional Round, 8.6

Given triangle $ABC$. Point $A_1$ is symmetric to vertex $A$ wrt line $BC$, and point $C_1$ is symmetric to vertex $C$ wrt line $AB$. Prove that if points $A_1$, $B$ and $C_1$ lie on the same line and $C_1B = 2A_1B$, then angle $\angle CA_1B$ is right.

2020 Thailand TST, 1

Tags: geometry , circumcircle , right angle , right triangle , tangent circle

Let $ABC$ be a triangle with circumcircle $\Gamma$. Let $\omega_0$ be a circle tangent to chord $AB$ and arc $ACB$. For each $i = 1, 2$, let $\omega_i$ be a circle tangent to $AB$ at $T_i$ , to $\omega_0$ at $S_i$ , and to arc $ACB$. Suppose $\omega_1 \ne \omega_2$. Prove that there is a circle passing through $S_1, S_2, T_1$, and $T_2$, and tangent to $\Gamma$ if and only if $\angle ACB = 90^o$. .

Geometry Mathley 2011-12, 10.1

Tags: right angle , incircle , circumcircle , tangent circle , geometry

Let $ABC$ be a triangle with two angles $B,C$ not having the same measure, $I$ be its incircle, $(O)$ its circumcircle. Circle $(O_b)$ touches $BA,BC$ and is internally tangent to $(O)$ at $B_1$. Circle $(O_c)$ touches $CA,CB$ and is internally tangent to $(O)$ at $C_1$. Let $S$ be the intersection of $BC$ and $B_1C_1$. Prove that $\angle AIS = 90^o$. Nguyễn Minh Hà

2007 Estonia Math Open Junior Contests, 7

Tags: geometry , angle bisector , square , right angle

The center of square $ABCD$ is $K$. The point $P$ is chosen such that $P \ne K$ and the angle $\angle APB$ is right . Prove that the line $PK$ bisects the angle between the lines $AP$ and $BP$.

2015 Dutch IMO TST, 1

Tags: right angle , equal angles , equal segments , geometry

In a quadrilateral $ABCD$ we have $\angle A = \angle C = 90^o$. Let $E$ be a point in the interior of $ABCD$. Let $M$ be the midpoint of $BE$. Prove that $\angle ADB = \angle EDC$ if and only if $|MA| = |MC|$.

Kvant 2024, M2809

Tags: conditional geometry , angle measure , construction , right angle , beautiful geometry

Given is a triangle $ABC$ and the points $M, P$ lie on the segments $AB, BC$, respectively, such that $AM=BC$ and $CP=BM$. If $AP$ and $CM$ meet at $O$ and $2\angle AOM=\angle ABC$, find the measure of $\angle ABC$.

2021 New Zealand MO, 1

Tags: geometry , right angle , equal segments

Let $ABCD$ be a convex quadrilateral such that $AB + BC = 2021$ and $AD = CD$. We are also given that $\angle ABC = \angle CDA = 90^o$. Determine the length of the diagonal $BD$.

2004 Germany Team Selection Test, 2

Tags: geometry , combinatorial geometry , polygon , extremal combinatorics , right angle , angle

Let $n \geq 5$ be a given integer. Determine the greatest integer $k$ for which there exists a polygon with $n$ vertices (convex or not, with non-selfintersecting boundary) having $k$ internal right angles. [i]Proposed by Juozas Juvencijus Macys, Lithuania[/i]

1985 Poland - Second Round, 6

Tags: geometry , 3d geometry , right angle

There are various points in space $ A, B, C_0, C_1, C_2 $, with $ |AC_i| = 2 |BC_i| $ for $ i = 0,1,2 $ and $ |C_1C_2|=\frac{4}{3}|AB| $. Prove that the angle $ C_1C_0C_2 $ is right and the points $ A, B, C_1, C_2 $ lie on one plane.

2003 IMO Shortlist, 3

Tags: geometry , combinatorial geometry , polygon , extremal combinatorics , right angle , angle

Let $n \geq 5$ be a given integer. Determine the greatest integer $k$ for which there exists a polygon with $n$ vertices (convex or not, with non-selfintersecting boundary) having $k$ internal right angles. [i]Proposed by Juozas Juvencijus Macys, Lithuania[/i]

2019 New Zealand MO, 2

Tags: geometry , angle bisector , right angle

Let $X$ be the intersection of the diagonals $AC$ and $BD$ of convex quadrilateral $ABCD$. Let $P$ be the intersection of lines $AB$ and $CD$, and let $Q$ be the intersection of lines $PX$ and $AD$. Suppose that $\angle ABX = \angle XCD = 90^o$. Prove that $QP$ is the angle bisector of $\angle BQC$.

2010 Abels Math Contest (Norwegian MO) Final, 1a

Tags: geometry , right angle , equal segments , angle

The point $P$ lies on the edge $AB$ of a quadrilateral $ABCD$. The angles $BAD, ABC$ and $CPD$ are right, and $AB = BC + AD$. Show that $BC = BP$ or $AD = BP$.

May Olympiad L2 - geometry, 2001.2

Tags: geometry , trapezoid , right angle

On the trapezoid $ABCD$ , side $DA$ is perpendicular to the bases $AB$ and $CD$ . The base $AB$ measures $45$, the base $CD$ measures $20$ and the $BC$ side measures $65$. Let $P$ on the $BC$ side such that $BP$ measures $45$ and $M$ is the midpoint of $DA$. Calculate the measure of the $PM$ segment.

2018 Dutch IMO TST, 3

Tags: right angle , equal segments , geometry

Let $ABC$ be an acute triangle, and let $D$ be the foot of the altitude through $A$. On $AD$, there are distinct points $E$ and $F$ such that $|AE| = |BE|$ and $|AF| =|CF|$. A point$ T \ne D$ satises $\angle BTE = \angle CTF = 90^o$. Show that $|TA|^2 =|TB| \cdot |TC|$.

2021 Dutch IMO TST, 2

Tags: right angle , centroid , right triangle , geometry

Let $ABC $be a right triangle with $\angle C = 90^o$ and let $D$ be the foot of the altitude from $C$. Let $E$ be the centroid of triangle $ACD$ and let $F$ be the centroid of triangle $BCD$. The point $P$ satisfies $\angle CEP = 90^o$ and $|CP| = |AP|$, while point $Q$ satisfies $\angle CFQ = 90^o$ and $|CQ| = |BQ|$. Prove that $PQ$ passes through the centroid of triangle $ABC$.

2018 Grand Duchy of Lithuania, 3

Tags: geometry , right angle , circumcenter

The altitudes $AD$ and $BE$ of an acute triangle $ABC$ intersect at point $H$. Let $F$ be the intersection of the line $AB$ and the line that is parallel to the side BC and goes through the circumcenter of $ABC$. Let $M$ be the midpoint of the segment $AH$. Prove that $\angle CMF = 90^o$

2003 Oral Moscow Geometry Olympiad, 2

Tags: ratio , geometry , right angle , equal segments , equal angles

In a convex quadrilateral $ABCD$, $\angle ABC = 90^o$ , $\angle BAC = \angle CAD$, $AC = AD, DH$ is the alltitude of the triangle $ACD$. In what ratio does the line $BH$ divide the segment $CD$?

1999 Poland - Second Round, 4

Tags: equal angles , right angle , circumcenter , geometry

Let $P$ be a point inside a triangle $ABC$ such that $\angle PAB = \angle PCA$ and $\angle PAC = \angle PBA$. If $O \ne P$ is the circumcenter of $\triangle ABC$, prove that $\angle APO$ is right.

Indonesia MO Shortlist - geometry, g9

Tags: right angle , arc midpoint , circles , geometry

Given two circles $\Gamma_1$ and $\Gamma_2$ which intersect at points $A$ and $B$. A line through $A$ intersects $\Gamma_1$ and $\Gamma_2$ at points $C$ and $D$, respectively. Let $M$ be the midpoint of arc $BC$ in $\Gamma_1$ ,which does not contains $A$, and $N$ is the midpoint of the arc $BD$ in $\Gamma_2$, which does not contain $A$. If $K$ is the midpoint of $CD$, prove that $\angle MKN = 90^o.$

Kyiv City MO Seniors Round2 2010+ geometry, 2018.11.2

Tags: circumcircle , geometry , right angle

In the quadrilateral $ABCD $, $AB = BC $, the point $K $ is the midpoint of the side $CD $, the rays $BK $ and $AD $ intersect at the point $M $ , the circumscribed circle $ \Delta ABM $ intersects the line $AC $ for the second time at the point $P $. Prove that $\angle BKP = 90 {} ^ \circ $. (Anton Trygub)

2001 May Olympiad, 2

Tags: geometry , trapezoid , right angle

On the trapezoid $ABCD$ , side $DA$ is perpendicular to the bases $AB$ and $CD$ . The base $AB$ measures $45$, the base $CD$ measures $20$ and the $BC$ side measures $65$. Let $P$ on the $BC$ side such that $BP$ measures $45$ and $M$ is the midpoint of $DA$. Calculate the measure of the $PM$ segment.

2020 China Team Selection Test, 2

Tags: right angle , equal segments , tangent circle , geometry

Given an isosceles triangle $\triangle ABC$, $AB=AC$. A line passes through $M$, the midpoint of $BC$, and intersects segment $AB$ and ray $CA$ at $D$ and $E$, respectively. Let $F$ be a point of $ME$ such that $EF=DM$, and $K$ be a point on $MD$. Let $\Gamma_1$ be the circle passes through $B,D,K$ and $\Gamma_2$ be the circle passes through $C,E,K$. $\Gamma_1$ and $\Gamma_2$ intersect again at $L \neq K$. Let $\omega_1$ and $\omega_2$ be the circumcircle of $\triangle LDE$ and $\triangle LKM$. Prove that, if $\omega_1$ and $\omega_2$ are symmetric wrt $L$, then $BF$ is perpendicular to $BC$.

2024 Bulgaria MO Regional Round, 12.1

Tags: incentre , geometry , conditional geometry , right angle , isosceles

Let $ABC$ be an acute triangle with midpoint $M$ of $AB$. The point $D$ lies on the segment $MB$ and $I_1, I_2$ denote the incenters of $\triangle ADC$ and $\triangle BDC$. Given that $\angle I_1MI_2=90^{\circ}$, show that $CA=CB$.

2017 Saudi Arabia Pre-TST + Training Tests, 9

Tags: geometry , circumcircle , symmedian , right angle

Let $ABC$ be a triangle inscribed in circle $(O)$, with its altitudes $BH_b, CH_c$ intersect at orthocenter $H$ ($H_b \in AC$, $H_c \in AB$). $H_bH_c$ meets $BC$ at $P$. Let $N$ be the midpoint of $AH, L$ be the orthogonal projection of $O$ on the symmedian with respect to angle $A$ of triangle $ABC$. Prove that $\angle NLP = 90^o$.

2017 Singapore Senior Math Olympiad, 2

Tags: geometry , right angle , perpendicular , cyclic quadrilateral , cyclic

In the cyclic quadrilateral $ABCD$, the sides $AB, DC$ meet at $Q$, the sides $AD,BC$ meet at $P, M$ is the midpoint of $BD$, If $\angle APQ=90^o$, prove that $PM$ is perpendicular to $AB$.